dpb56083 2018-10-05 09:32
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PHP当函数作为参数传递时,如何获取函数参数

I want to do something like this:

function func($callback) {
    $result = $callback(???);    // $callback must be called here
    //...
}

//...
func(function(['foo' => 'buu']) {
    $a = func_get_arg(0);

    // do something with $a and return something...    
    return $something;       
})

It is possible in php?

I can do something like below, but this is not what I want to do:

function func($args, $callback) {
    $result = $callback($args);
    //...
}

func(['foo' => 'boo'], function($args) {
     $a = $args; // etc.
})
  • 写回答

1条回答 默认 最新

  • douguang9014 2018-10-05 09:39
    关注

    I use anonymous function this ways :

    $mySuperFunction = function($arg)
    {
        echo ("Hello " . $arg);
    };
    
    function AnotherFunction($args)
    {
        echo ("Another hello " . $args);
    }
    
    function WrappingAnonymous($callback, $args)
    {
        $callback($args);
    }
    
    function WrappingAnonymousWithoutArgs($callback)
    {
        $callback();
    }
    
    WrappingAnonymous($mySuperFunction, "World");
    WrappingAnonymous("AnotherFunction", "World");
    WrappingAnonymous(function($someArgs)
    {
        echo "Yet another Hello " . $someArgs;
    }, "World");
    
    
    WrappingAnonymousWithoutArgs(function($someArgs = "World")
    {
        echo "Now, a 4th other Hello " . $someArgs;
    });
    

    Outputs :

    Hello World

    Another hello World

    Yet another Hello World

    Now, a 4th other Hello World

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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